《2025年湘岳假期暑假作业八年级数学人教版》

18. 阅读下面的材料并解决问题。
$\frac {1}{\sqrt {2} + 1} = \frac {\sqrt {2} - 1}{(\sqrt {2} + 1)(\sqrt {2} - 1)} = \sqrt {2} - 1；$
$\frac {1}{\sqrt {3} + \sqrt {2}} = \frac {\sqrt {3} - \sqrt {2}}{(\sqrt {3} + \sqrt {2})(\sqrt {3} - \sqrt {2})} = \sqrt {3} - \sqrt {2}；$
$\frac {1}{2 + \sqrt {3}} = \frac {2 - \sqrt {3}}{(2 + \sqrt {3})(2 - \sqrt {3})} = 2 - \sqrt {3}；$
…
(1)观察上式并填空：$\frac {1}{\sqrt {7} + \sqrt {6}} = $______$\sqrt{7}-\sqrt{6}$。
(2)观察上述规律并猜想：当n是正整数时$,\frac {1}{\sqrt {n + 1} + \sqrt {n}} = $______$\sqrt{n + 1}-\sqrt{n}$(用含n的式子表示)。
(3)请利用(2)的结论计算：
$(\frac {1}{\sqrt {2} + 1} + \frac {1}{\sqrt {3} + \sqrt {2}} + … + \frac {1}{\sqrt {2022} + \sqrt {2021}} + \frac {1}{\sqrt {2023} + \sqrt {2022}})×(\sqrt {2023} +1)$。
解：原式$=(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\cdots+\sqrt{2022}-\sqrt{2021}+\sqrt{2023}-\sqrt{2022})×(\sqrt{2023}+1)$
$=(\sqrt{2023}-1)×(\sqrt{2023}+1)$
$=2023 - 1$
$=2022$

答案： 解：
(1) $\sqrt{7}-\sqrt{6}$
(2) $\sqrt{n + 1}-\sqrt{n}$
(3) 原式 $=(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\cdots+\sqrt{2022}-\sqrt{2021}+\sqrt{2023}-\sqrt{2022})×(\sqrt{2023}+1)$
$=(\sqrt{2023}-1)×(\sqrt{2023}+1)$
$=2023 - 1$
$=2022$.